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42
max
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�Pmech = 30 000 M · n
PJ = R · Imot2
Pel = Umot · Imot
Key information May 2014 edition / subject to change
maxon DC motor and maxon EC motorKey information
The motor as an energy converter
The electrical motor converts electrical power Pel (current Imot and voltage Umot) into mechanical power Pmech (speed n and torque M). The losses that arise are divided into frictional losses, attributable to Pmech and in Joule power losses PJ of the winding (resistance R). Iron losses do not occur in the coreless maxon DC motors. In maxon EC motors, they are treated formally like an additional friction torque. The power balance can therefore be formulated as:
Pel = Pmech + PJ
The detailed result is as follows�Umot · Imot = 30 000 n · M + R · Imot
2
Electromechanical motor constantsThe geometric arrangement of the magnetic circuit and winding defines in detail how the motor converts the electrical input power (current, voltage) into mechanical output power (speed, torque). Two important characteristic values of this energy conversion are the speed constant kn and the torque constant kM. The speed constant combines the speed n with the voltage induced in the winding Uind (= EMF). Uind is proportional to the speed; the following applies:
n = kn · Uind
Similarly, the torque constant links the mechanical torque M with the elec-trical current Imot.
M = kM · Imot
The main point of this proportionality is that torque and current are equivalent for the maxon motor.The current axis in the motor diagrams is therefore shown as parallel to the torque axis as well.
See also: Technology – short and to the point, explanation of the motor
Motor diagrams
A diagram can be drawn for every maxon DC and EC motor, from which key motor data can be taken. Although tolerances and temperature influ-ences are not taken into consideration, the values are sufficient for a first estimation in most applications. In the diagram, speed n, current Imot, power output P2 and efficiency h are applied as a function of torque M at constant voltage Umot.
Speed-torque lineThis curve describes the mechanical behavior of the motor at a constant voltage Umot: – Speed decreases linearly with increasing torque.– The faster the motor turns, the less torque it can provide.The curve can be described with the help of the two end points, no load speed n0 and stall torque MH (cf. lines 2 and 7 in the motor data).DC motors can be operated at any voltage. No load speed and stall torque change proportionally to the applied voltage. This is equivalent to a parallel shift of the speed-torque line in the diagram. Between the no load speed and voltage, the following proportionality applies in good approximation
n0 ≈ kn · Umot
where kn is the speed constant (line 13 of the motor data).
Independent of the voltage, the speed-torque line is described most practi-cally by the slope or gradient of the curve (line 14 of the motor data).
n0 = MH
ΔnΔM
Derivation of the speed-torque lineThe following occurs if one replaces current Imot with torque M using the torque constant in the detailed power balance:
Umot · n · M + R ·� = 30 000MkM
MkM
2
Transformed and taking account of the close relationship of kM and kn, an equation is produced of a straight line between speed n and torque M.
n = kn · Umot −30 000
�R
k M2· · M
or with the gradient and the no load speed n0
· MΔnΔMn = n0 −
UnitsIn all formulas, the variables are to be used in the units according to the catalog (cf. physical variables and their units on page 48).
The following applies in particular:– All torques in mNm– All currents in A (even no load currents)– Speeds (rpm) instead of angular velocity (rad/s)
Motor constantsSpeed constant kn and torque constant kM are not independent of one another. The following applies:
The speed constant is also called specific speed. Specific voltage, generator or voltage constants are mainly the reciprocal value of the speed constant and describe the voltage induced in the motor per speed. The torque constant is also called specific torque. The recipro-cal value is called specific current or current constant.
Speed n
Torque M
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IN
MN
nN
U = UN
n0
IA
n0
MH
May 2014 edition / subject to change Key information
The speed-torque gradient is one of the most informative pieces of data and allows direct comparison between different motors. The smaller the speed-torque gradient, the less sensitive the speed reacts to torque (load) changes and the stronger the motor. With the maxon motor, the speed-torque gradient within the winding series of a motor type (i.e. on one catalog page) remains practically constant.
Current gradientThe equivalence of current to torque is shown by an axis parallel to the torque: more current flowing through the motor produces more torque. The current scale is determined by the two points no load current I0 and starting current IA (lines 3 and 8 of motor data). The no load current is equivalent to the friction torque MR, that describes the internal friction in the bearings and commutation system.
MR = kM · I0
In the maxon EC motor, there are strong, speed dependent iron losses in the stator iron stack instead of friction losses in the commutation system.
The motors develop the highest torque when starting. It is many times greater than the normal operating torque, so the current uptake is the greatest as well.The following applies for the stall torque MH and starting current IA
MH = kM · IA
Efficiency curveThe efficiency h describes the relationship of mechanical power delivered to electrical power consumed.
η =�
30 000n · (M − MR)
Umot · Imot·
One can see that at constant applied voltage U and due to the proportion-ality of torque and current, the efficiency increases with increasing speed (decreasing torque). At low torques, friction losses become increasingly significant and efficiency rapidly approaches zero. Maximum efficiency (line 9 of motor data) is calculated using the starting current and no load current and is dependent on voltage.
ηmax = 1 −I0
IA
2
Maximum efficiency and maximum output power do not occur at the same torque.
Rated operating point
The rated operating point is an ideal operating point for the motor and derives from operation at nominal voltage UN (line 1 of motor data) and nominal current IN (line 6). The nominal torque MN produced (line 5) in this operating point follows from the equivalence of torque and current.
MN kM · (IN – I0)
Nominal speed nN (line 4) is reached in line with the speed gradient. The choice of nominal voltage follows from considerations of where the maximum no load speed should be. The nominal current derives from the motor’s thermally maximum permissible continuous current.
Speed n
Torque M
Current I
Speed n
Torque M
Current I
Torque M
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10 20 40
20000
15000
10000
5000
0.4 0.8 1.2
25000
30
5
4
3
2
1
010 20 30 40 50 60 70 80 90 tON%
TION / IN
Key information May 2014 edition / subject to change
Operating range diagram
Motor diagrams, operating ranges
The catalogue contains a diagram of every maxon DC and EC motor type that shows the operating ranges of the different winding types using a typi-cal motor.
Permanent operating rangeThe two criteria “maximum continuous torque” and “maximum permis- sible speed” limit the continuous operating range. Operating points within this range are not critical thermally and do not generally cause increased wear of the commutation system.
Short-term operating rangeThe motor may only be loaded with the maximum continuous current for thermal reasons. However, temporary higher currents (torques) are al-lowed. As long as the winding temperature is below the critical value, the winding will not be damaged. Phases with increased currents are time limited. A measure of how long the temporary overload can last is provided by the thermal time constant of the winding (line 19 of the motor data). The magnitude of the times with overload ranges from several seconds for the smallest motors (6 mm to 13 mm diameter) up to roughly one minute for the largest (60 mm to 90 mm diameter). The calculation of the exact overload time is heavily dependent on the motor current and the ro-tor’s starting temperature.
Maximum continuous current, maximum continuous torqueThe Jule power losses heat up the winding. The heat produced must be able to dissipate and the maximum rotor temperature (line 22 of the mo-tor data) should not be exceeded. This results in a maximum continuous current, at which the maximum winding temperature is attained under standard conditions (25°C ambient temperature, no heat dissipation via the flange, free air circulation). Higher motor currents cause excessive winding temperatures. The nominal current is selected so that it corresponds to this maximum permissible constant current. It depends heavily on the winding. These thin wire windings have lower nominal current levels than thick ones. With very low resistive windings, the brush system’s capacity can further limit the permissible constant current. With graphite brush motors, friction losses increase sharply at higher speeds. With EC motors, eddy current losses increase in the return as speed increases and produce additional heat. The maximum permissible continuous current decreases at faster speeds accordingly. The nominal torque allocated to the nominal current is almost constant within a motor type’s winding range and represents a characteristic size of the motor type.
The maximum permissible speedfor DC motors is primarily limited by the commutation system. The commutator and brushes wear more rapidly at very high speeds. The reasons are:– Increased mechanical wear because of the large traveled path of
the commutator – Increased electro-erosion because of brush vibration and spark
formation.
A further reason for limiting the speed is the rotor’s residual mechanical imbalance which shortens the service life of the bearings. Higher speeds than the limit speed nmax (line 23) are possible, however, they are “paid for” by a reduced service life expectancy. The maximum permissible speed for the EC motor is calculated based on service life considerations of the ball bearings (at least 20 000 hours) at the maximum residual imbalance and bearing load.
Maximum winding temperatureThe motor current causes the winding to heat up due to the winding’s resistance. To prevent the motor from overheating, this heat must dissipate to the environment via the stator. The coreless winding is the thermally critical point. The maximum rotor temperature must not be exceeded, even temporarily. With graphite brush motors and EC motors which tend to have higher current loads, the maximum rotor temperature is 125°C (in individual cases up to 155°C). Motors with precious metal commutators only allow lower current loads, so that the rotor temperatures must not exceed 85°C. Favourable mounting conditions, such as good air circulation or cooling plates, can significantly lower temperatures.
ON Motor in operationOFF Motor stationary ION Max. peak current IN Max. permissible continuous current (line 6)tON ON time [s], should not exeed tw (line 19) T Cycle time tON + tOFF [s]tON% Duty cycle as percentage of cycle time. The motor may be overloaded by the relationship ION / IN at X % of the total cycle time.
Ion = IN T
tON
Time
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n n
M
n n
M
U>UN
U<UN
U=UN
May 2014 edition / subject to change Key information
maxon flat motor
Multipole EC motors, such as maxon flat motors, require a greater number of commutation steps for a motor revolution (6 x number of pole pairs). Due to the wound stator teeth they have a higher terminal inductance than mo-tors with an ironless winding. As a result at higher speed, the current can-not develop fully during the correspondingly short commutation intervals. Therefore, the apparent torque produced is lower. Current is also fed back into the controller’s power stage. As a result, motor behaviour deviates from the ideal linear speed-torque gradient. The apparent speed-torque gradient depends on voltage and speed: The gradient is steeper at higher speeds.Mostly, flat motors are operated in the continuous operation range where the achievable speed-torque gradient at nominal voltage can be approxi-mated by a straight line between no load speed and nominal operating point. The achievable speed-torque gradient is approximately.
≈Δn
ΔMn0 − nN
MN
Acceleration
In accordance with the electrical boundary conditions (power supply, control, battery), a distinction is principally made between two different starting processes:– Start at constant voltage (without current limitation)– Start at constant current (with current limitation)
Start under constant currentA current limit always means that the motor can only deliver a limited torque. In the speed-torque diagram, the speed increases on a vertical line with a constant torque. Acceleration is also constant, thus simplifying the calculation. Start at constant current is usually found in applications with servo amplifiers, where acceleration torques are limited by the amplifier’s peak current.
– Angular acceleration a (in rad / s2) at constant current I or constant torque M with an additional load of inertia JL:
kM · Imot
JR + JL
MJR + JL
α = 104 · = 104 ·
– Run-up time Dt (in ms) at a speed change Dn with an additional load inertia JL:
�300
JR + JL
kM · ImotΔt = · Δn ·
(all variables in units according to the catalog)
Start with constant terminal voltageHere, the speed increases from the stall torque along the speed- torque line. The greatest torque and thus the greatest acceleration is effective at the start. The faster the motor turns, the lower the accel-eration. The speed increases more slowly. This exponentially flattening increase is described by the mechanical time constant tm (line 15 of the motor data). After this time, the rotor at the free shaft end has at-tained 63% of the no load speed. After roughly three mechanical time constants, the rotor has almost reached the no load speed.
– Mechanical time constant tm (in ms) of the unloaded motor:
JR · RkM
2τm = 100 ·
– Mechanical time constants tm’ (in ms) with an additional load inertia JL:
τm' = 100 · 1 +JR · R
kM2
JL
JR
– Maximum angular acceleration amax (in rad / s2) of the unloaded motor:
MH
JRαmax = 104 ·
– Maximum angular acceleration amax (in rad / s2) with an additional load inertia JL:
MH
JR + JLαmax = 104 ·
– Run-up time (in ms) at constant voltage up to the operating point (ML , nL ):
Δt = τm' · In
ML + MR
MH1 – · n0
1 –ML + MR
MH
· n0 – nL
l = constant
Time
U = constant
Time
Speed n
Torque M
UN ideal
Nominal operating point
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Tolerances
Tolerances must be considered in critical ranges. The possible deviations of the mechanical dimensions can be found in the overview drawings. The motor data are average values: the adjacent diagram shows the effect of tolerances on the curve characteristics. They are mainly caused by dif-ferences in the magnetic field strength and in wire resistance, and not so much by mechanical influences. The changes are heavily exaggerated in the diagram and are simplified to improve understanding. It is clear, how-ever, that in the motor’s actual operating range, the tolerance range is more limited than at start or at no load. Our computer sheets contain all detailed specifications.
CalibratingThe tolerances can be limited by controlled de-magnetization of the motors. Motor data can be accurately specified down to 1 to 3%. However, the motor characteristic values lie in the lower portion of the standard tolerance range.
Thermal behavior
The Joule power losses PJ in the winding determine heating of the motor. This heat energy must be dissipated via the surfaces of the winding and motor. The increase DTW of the winding temperature TW with regard to the ambient temperature arises from heat losses PJ and thermal resistances Rth1 and Rth2.
TW − TU = DTW = (Rth1 + Rth2) · PJ
Here, thermal resistance Rth1 relates to the heat transfer between the wind-ing and the stator (magnetic return and magnet), whereas Rth2 describes the heat transfer from the housing to the environment. Mounting the motor on a heat dissipating chassis noticeably lowers thermal resistance Rth2. The values specified in the data sheets for thermal resistances and the maxi-mum continuous current were determined in a series of tests, in which the motor was end-mounted onto a vertical plastic plate. The modified thermal resistance Rth2 that occurs in a particular application must be determined using original installation and ambient conditions. Thermal resistance Rth2 on motors with metal flanges decreases by up to 80% if the motor is coupled to a good heat-conducting (e.g. metallic) retainer.
The heating runs at different rates for the winding and stator due to the different masses. After switching on the current, the winding heats up first (with time constants from several seconds to half a minute). The stator reacts much slower, with time constants ranging from 1 to 30 minutes depending on motor size. A thermal balance is gradually established. The temperature difference of the winding compared to the ambient tempera-ture can be determined with the value of the current I (or in intermittent operation with the effective value of the current I = IRMS ).
(Rth1 + Rth2) · R · I mot2
1– αCu · (Rth1 + Rth2) · R · I mot2ΔTW =
Here, electrical resistance R must be applied at the actual ambient temperature.
Influence of temperatureAn increased motor temperature affects winding resistance and mag-netic characteristic values.
Winding resistance increases linearly according to the thermal resistance coefficient for copper (αCu = 0.0039):
RT = R25 · (1 + αCu (T −25°C ))
Example: a winding temperature of 75°C causes the winding resist- ance to increase by nearly 20%.
The magnet becomes weaker at higher temperatures. The reduction is 1 to 10% at 75°C depending on the magnet material.
The most important consequence of increased motor temperature is that the speed curve becomes steeper which reduces the stall torque. The changed stall torque can be calculated in first approximation from the voltage and increased winding resistance:
Umot
RTMHT = kM · IAT = kM ·
Tolerance field presentationfor maxon motors
Tolerance forstarting current
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n
M
Speed-torque line high enoughfor the required load speed
U = constant
Speed-torqueline too low for therequired load speed
Motor selection
The drive requirements must be defined before proceeding to motor selec-tion.– How fast and at which torques does the load move?– How long do the individual load phases last?– What accelerations take place?– How great are the mass inertias?Often the drive is indirect, this means that there is a mechanical transfor-mation of the motor output power using belts, gears, screws and the like. The drive parameters, therefore, are to be calculated to the motor shaft. Additional steps for gear selection are listed below.
Furthermore, the power supply requirements need to be checked.– Which maximum voltage is available at the motor terminals?– Which limitations apply with regard to current?The current and voltage of motors supplied with batteries or solar cells are very limited. In the case of control of the unit via a servo amplifier, the amplifier’s maximum current is often an important limit.
Selection of motor typesThe possible motor types are selected using the required torque. On the one hand, the peak torque, Mmax, is to be taken into consideration and on the other, the effective torque MRMS. Continuous operation is characterized by a single operating or load point (ML, nL). The motor types in question must have a nominal torque (= max. continuous torque) MN that is greater than load torque ML.
MN > ML
In work cycles, such as start/stop operation, the motor‘s nominal torque must be greater than the effective load torque (quadratically averaged). This prevents the motor from overheating.
MN > MRMS
The stall torque of the selected motor should usually exceed the emerging load peak torque.
MH > Mmax
Selection of the winding: electric requirementIn selecting the winding, it must be ensured that the voltage applied directly to the motor is sufficient for attaining the required speed in all operating points.
Unregulated operationIn applications with only one operating point, this is often achieved with a fixed voltage U. A winding is sought with a speed-torque line that passes through the operating point at the specified voltage. The calculation uses the fact that all motors of a type feature practically the same speed-torque gradient. A target no load speed n0,theor is calculated from operating point (nL, ML).
ΔnΔMn0, theor = nL + ML
This target no load speed must be achieved with the existing voltage U, which defines the target speed constant.
n0, theor
Umotkn, theor =
Those windings whose kn is as close to kn, theor as possible, will approximate the operating point the best at the specified voltage. A somewhat larger speed constant results in a somewhat higher speed, a smaller speed con-stant results in a lower one. The variation of the voltage adjusts the speed to the required value, a principle that servo amplifiers also use.
The motor current Imot is calculated using the torque constant kM of the selected winding and the load torque ML.
ML
kMImot =
Advices for evaluating the requirements:Often the load points (especially the torque) are not known or are difficult to determine. In such cases you can operate your device with a measuring motor roughly estimated according to size and power. Vary the voltage until the desired operating points and motion sequences have been achieved. Measure the voltage and current flow. Using these specifications and the part number of the measuring motor, our engineers can often specify the suitable motor for your application.
Additional optimization criteria are, for example:– Mass to be accelerated (type, mass inertia)– Type of operation (continuous, intermittent, reversing)– Ambient conditions (temperature, humidity, medium)– Power supply, battery
When selecting the motor type, other constraints also play a major role: – What maximum length should the drive unit have, including gear and
encoder? – What diameter? – What service life is expected from the motor and which
commutation system should be used?– Precious metal commutation for continuous operation at
low currents (rule of thumb for longest service life: up to approx. 50% of IN ).
– Graphite commutation for high continuous currents (rule of thumb: 50% to approx. 75% of IN ) and frequent current peaks (start/stop operation, reversing operation).
– Electronic commutation for highest speeds and longest service life.– How great are the forces on the shaft, do ball bearings have to be used
or are less expensive sintered bearings sufficient?
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n
M
n
0.5 2.5 3.0 3.7
Example for motor/gear selection
A drive should move cyclically according to the following speed diagram.
The inertia of the load to be accelerated JL is 140 000 gcm2. The constant coefficient is approximately 300 mNm. The 4-quadrant operation allows controlled and dynamic motor operation and brake operation in two direc-tions of rotation (all 4 quadrants). The power supply unit delivers max. 3 A and 24 V.
Calculation of load dataThe torque required for acceleration and braking are calculated as follows (motor and gearhead inertia omitted):
πMα = JL � 30ΔnΔt = 0.014 � π
30600.5� = 0.176 Nm = 176 mNm
Together with the friction torque, the following torques result for the different phases of motion.– Acceleration phase (duration 0.5 s) 476 mNm – Constant speed (duration 2 s) 300 mNm – Braking (friction brakes
with 300 mNm) (duration 0.5 s) 124 mNm – Standstill (duration 0.7 s) 0 mNm
Peak torque occurs during acceleration. The RMS determined torque of the entire work cycle is
t1 · M2
1 + t2 · M2
2 + t3 · M2
3 + t4 · M2
4MRMS = ttot
0.5 · 4762 + 2 · 3002 + 0.5 · 1242 + 0.7 · 0=
3.7≈ 285 mNm
The maximum speed (60 rpm) occurs at the end of the acceleration phase at maximum torque (463 mNm). Thus, the peak mechanical power is:
πPmax = Mmax � 30 nmax = 0.476 � π30 � 60 ≈ 3 W
Regulated servo drivesIn work cycles, all operating points must lie beneath the curve at a maxi-mum voltage Umax. Mathematically, this means that the following must apply for all operating points (nL, ML):
ΔnΔMkn · Umax = n0 > nL + ML
When using servo amplifiers, a voltage drop occurs at the power stage, so that the effective voltage applied to the motor is lower. This must be taken into consideration when determining the maximum supply voltage Umax. It is recommended that a regulating reserve of some 20% be included, so that regulation is even ensured with an unfavorable tolerance situation of motor, load, amplifier and supply voltage. Finally, the average current load and peak current are calculated ensuring that the servo amplifier used can deliver these currents. In some cases, a higher resistance winding must be selected, so that the currents are lower. However, the required voltage is then increased.
Physical variables and their units SI Catalogi Gear reduction* Imot Motor current A A, mAIA Stall current* A A, mAI0 No load current* A mAIRMS RMS determined current A A, mAIN Nominal current* A A, mAJR Moment of inertia of the rotor* kgm2 gcm2
JL Moment of inertia of the load kgm2 gcm2
kM Torque constant* Nm/A mNm/Akn Speed constant* rpm/VM (Motor) torque Nm mNmML Load torque Nm mNmMH Stall torque* Nm mNmMmot Motor torque Nm mNmMR Moment of friction Nm mNmMRMS RMS determined torque Nm mNmMN Nominal torque Nm mNmMN,G Max. torque of gear* Nm Nmn Speed rpmnL Operating speed of the load rpmnmax Limit speed of motor* rpmnmax,G Limit speed of gear* rpmnmot Motor speed rpmn0 No load speed* rpmPel Electrical power W WPJ Joule power loss W WPmech Mechanical power W WR Terminal resistance W WR25 Resistance at 25°C* W WRT Resistance at temperature T W WRth1 Heat resistance winding housing* K/WRth2 Heat resistance housing/air* K/Wt Time s sT Temperature K °CTmax Max. winding temperature* K °CTU Ambient temperature K °CTW Winding temperature K °CUmot Motor voltage V VUind Induced voltage (EMF) V VUmax Max. supplied voltage V VUN Nominal voltage* V VaCu Resistance coefficient of Cu = 0.0039amax Maximum angle acceleration rad/s2
Dn/DM Curve gradient* rpm/mNmDTW Temperature difference winding/ambient K KDt Run up time s msh (Motor) efficiency %hG (Gear) efficiency* %hmax Maximum efficiency* %tm Mechanical time constant* s mstS Therm. time constant of the motor* s stW Therm. time constant of the winding* s s(*Specified in the motor or gear data)
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n[rpm]
9.3 mNm, 5040 rpm
Gear selectionA gear is required with a maximum continuous torque of at least 0.28 Nm and an intermittent torque of at least 0.47 Nm. This requirement is fulfilled, for example, by a planetary gear with 22 mm diameter (metal version GB 22 A).The recommended input speed of 6000 rpm allows a maximum reduction of:
imax = nmax, G
nB= 100:1= 6000
60
We select the three-stage gear with the next smallst reduction of 84 : 1 (stock program). Efficiency is max. 59%.
Motor type selectionSpeed and torque are calculated to the motor shaft
nmot = i · nL = 84 · 60 = 5040 rpm
MRMSMmot, RMS = i � ɳ = 28584� 0.59 ≈ 5.8 mNm
MmaxMmot, max = i � ɳ = 47684� 0.59 ≈ 9.6 mNm
The possible motors, which match the selected gears in accordance with the maxon modular system, are summarized in the table opposite. The table contains only DC motors with graphite commutation, which are better suited for start-stop operation, as well as brushless EC motors.
Selection falls on an A-max 22, 6 W, which demonstrates a sufficiently high continuous torque. The motor should have a torque reserve so that it can even function with a somewhat unfavorable gear efficiency. The additional torque requirement during acceleration can easily be delivered by the mo-tor. The temporary peak torque is not even twice as high as the continuous torque of the motor.
Selection of the windingThe motor type A-max 22, 6 W has an average speed-torque gradient of some 450 rpm/mNm. However, it should be noted that the two lowest resistance windings have a somewhat steeper gradient. The desired no load speed is calculated as follows:
Δnn0, theor = nmot + ΔM � Mmax = 5040 + 450 � 9.6 = 9360 rpm
The extreme operating point should of course be used in the calculation (max. speed and max. torque), since the speed-torque line of the winding must run above all operating points in the speed / torque diagram.This target no load speed must be achieved with the maximum voltage U = 24 V supplied by the control (ESCON 36/2), which defines the mini-mum target speed constant kn, theor of the motor.
n0, theorkn, theor = Umot
= = 390936024
rpmV
According to the calculations, the selection of the motor is 110163, which with its speed constant of 558 rpm/V has a speed control reserve of over 20%. This means that unfavorable tolerances are not a problem. The higher speed constant of the winding compared to the calculated value means that the motor runs faster at 24 V than required, which can be compensat-ed with the controller. This motor also has a second shaft end for mounting an encoder. The torque constant of this winding is 17.1 mNm/A. Therefore the maxi-mum torque corresponds to a peak current of:
MmaxImax = kM+ I0 = + 0.029 = 0.6 A 9.6
17.1
This current is smaller than the maximum current (4 A) of the controller and the power supply unit (3 A).
Thus, a gear motor has been found that fulfils the requirements (torque and speed) and can be operated by the controller provided.
Motor MN SuitabilityA-max 22, 6 W ≈ 6.9 mNm goodA-max 19, 2.5 W ≈ 3.8 mNm too weakRE-max 21, 6 W ≈ 6.8 mNm goodEC 16, 30 W ≈ 8.5 mNm goodEC 16, 60 W ≈ 17 mNm too strongEC 20 flat, 3 W ≈ 3-4 mNm too weakEC 20 flat, 5 W ≈ 7.5 mNm goodEC 20 flat, 5 W, iE. ≈ 7.5 mNm good, possible alternative with
integrated speed controller, no ESCON control necessary
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B A oz-in-s-1 oz-in-min-1 in-lbf-s-1 ft-lbf-s-1 W = N · ms-1 mW kpm s-1 mNm min-1
W = N · ms-1 7.06 · 10-3 1.17 · 10-4 0.113 1.356 1 1 · 10-3 9.807 2π⁄60000
mW 7.06 0.117 112.9 1.356 · 103 1 · 103 1 9.807 · 103 2π⁄60
oz-in-s-1 1 1/60 16 192 141.6 0.142 1.39 · 103 2.36 · 10-3
ft-lbf-s-1 1⁄1921⁄11520
1⁄12 1 0.737 0.737 · 10-3 7.233 1.23 · 10-5
kpm s-1 7.20 · 10-4 1.2 · 10-5 1.15 · 10-2 0.138 0.102 0.102 · 10-3 1 1.70 · 10-6
B A oz-in ft-lbf Nm = Ws Ncm mNm kpm pcm
Nm 7.06 · 10-3 1.356 1 1 · 10-2 1 · 10-3 9.807 9.807 · 10-5
mNm 7.06 1.356 · 103 1 · 103 10 1 9.807 · 103 9.807 ·10-2
kpm 7.20 · 10-4 0.138 0.102 0.102 · 10-2 0.102 · 10-3 1 1 · 10-5
oz-in 1 192 141.6 1.416 0.142 1.39 · 103 1.39 · 10-2
ft-lbf 1⁄192 1 0.737 0.737 · 10-2 0.737 · 10-3 7.233 7.233 · 10-5
B A oz-in2 oz-in-s2 lb-in2 lb-in-s2 Nms2=kgm2 mNm s2 gcm2 kpm s2
g cm2 182.9 7.06 · 104 2.93 · 103 1.13 · 106 1 · 107 1 · 104 1 9.807 · 107
kgm2=Nms2 1.83 · 10-5 7.06 · 10-3 2.93 · 10-4 0.113 1 1 · 10-3 1 · 10-7 9.807
oz-in2 1 386.08 16 6.18 · 103 5.46 · 104 54.6 5.46 ·10-3 5.35 · 105
lb-in2 1⁄16 24.130 1 386.08 3.41 · 103 3.41 3.41 ·10-4 3.35 · 104
B A oz lb gr (grain) kg g B A oz lbf N kp p
kg 28.35 ·10-3 0.454 64.79·10-6 1 1 · 10-3 N 0.278 4.448 1 9.807 9.807·10-3
g 28.35 0.454 · 103 64.79·10-3 1 · 103 1 kp 0.028 0.454 0.102 1 1 · 10-3
oz 1 16 2.28 · 10-3 35.27 35.27 · 103 oz 1 16 3.600 35.27 35.27·10-3
lb 1⁄16 1 1⁄7000 2.205 2.205 · 103 lbf 1⁄16 1 0.225 2.205 2.205·10-3
gr (grain) 437.5 7000 1 15.43 · 10315.43 · 106 pdl 2.011 32.17 7.233 70.93 70.93·10-3
B A in ft yd Mil m cm mm m
m 25.4 · 10-3 0.305 0.914 25.4 · 10-6 1 0.01 1 · 10-3 1 · 10-6
cm 2.54 30.5 91.4 25.4 · 10-4 1 · 102 1 0.1 1 · 10-4
mm 25.4 305 914 25.4 · 10-3 1 · 103 10 1 1 · 10-3
in 1 12 36 1 · 10-3 39.37 0.394 3.94 · 10-2 3.94 · 10-5
ft 1⁄12 1 3 1⁄12 · 10-3 3.281 3.281 · 10-2 3.281 · 10-3 3.281 · 10-6
B A s-1 = Hz min-1 rad s-1 B A min-2 s-2 rad s-2 min-1 s-1
rad s-1 2p p ⁄30 1 s-2 1⁄3600 1 1⁄2p 1⁄60
min-1 1⁄60 1 30⁄p rad s-2 p ⁄1800 2p 1 p ⁄30
B A in-s-1 in-min-1 ft-s-1 ft-min-1 m s-1 cm s-1 mm s-1 m min-1
m s-1 2.54 · 10-2 4.23 · 10-4 0.305 5.08 · 10-3 1 1 · 10-2 1 · 10-3 1⁄60
in-s-1 1 60 12 720 39.37 39.37 · 10-2 39.37 · 10-3 0.656
ft-s-1 1⁄12 5 1 60 3.281 3.281 · 10-2 3.281 · 10-3 5.46 · 10-2
B A
(°F -305.15) / 1.8 + 273.15 1
(°F -32) / 1.8 1 -273.15
1 1.8°C + 32 1.8 K + 305.15
Key information May 2014 edition / subject to change
maxon Conversion Tables
General InformationQuantities and their basic units in the International System of Measurements (SI)
Quantity Basic- unit
Sign
LengthMassTimeElectrical currentThermodynamicTemperature
MeterKilogramSecondAmpere
Kelvin
mkgsA
K
Conversion ExampleA known unitB unit sought
known: multiply by sought:oz-in 7.06 mNm
Factors used for …
… conversions:1 oz = 2.834952313 · 10-2 kg1 in = 2.54 · 10-2 m
… gravitational acceleration:g = 9.80665 m s-2
= 386.08858 in s-2
… derived units:1 yd = 3 ft = 36 in1 lb = 16 oz = 7000 gr (grains)1 kp = 1 kg · 9.80665 ms-2
1 N = 1 kgms-2
1 W = 1 Nms-1 = 1 kgm2s-3
1 J = 1 Nm = 1 Ws
Decimal multiples and fractions of units
Prefix Abbre-viation
Multiply Prefix Abbre-viation
Multiply
Deka . . da 101 Dezi . . d 10-1
Hekto . . h 102 Zenti . . c 10-2
Kilo . . k 103 Milli . . m 10-3
Mega . . M 106 Mikro . . m 10-6
Giga . . G 109 Nano . . n 10-9
Tera . . T 1012 Piko . . p 10-12
Arc definition
Units used in this brochure
Power P [W]
Torque M [Nm]
Moment of Inertia J [kg m2]
Mass m [kg] Force F [N]
Length l [m]
Angular Velocity w [s-1] Angular Acceleration a [s-2]
Linear Velocity v [m s-1]
Temperature T [K]
° Fahrenheit ° Celsius = Centigrade Kelvin
Kelvin
° Celsius
° Fahrenheit
rpm
rpm